I can't figure out how to prove sin^2 + cosin^2 = 1 Can someone help me out?

Answer 1

The equation sin^2 + cosin^2 = 1 is really the Pythagorean theorem in disguise. Take a look.

First remember that sin=opp/hyp and cos=adj/hyp.

Squaring sin gives you opp^2/hyp^2.

Squaring cos gives you adj^2/hyp^2.

So now we have opp^2/hyp^2+adj^2/hyp^2=1

We have a common denominator of hyp^2, so we can rewrite the equation as (opp^2+adj^2)/hyp^2=1

Now just multiply both sides by hyp^2 to arrive at the Pythagorean theorem. opp^2+adj^2=hyp^2

If you need to continue and prove the Pythagorean theorem to be true, check out this website

http://www.apronus.com/geometry/pythagoras.htm

There is a proof of the theorem on there, although many many more do exist.

Answer 2

The proof requires the use of a diagram. Draw out a right triangle with hypotenuse equal to 1 unit in length. By definition, when the hypotenuse is equal to length 1, the sine is the leg of the triangle opposite the angle. Find this side and label it x. Now, the other leg, which is adjacent to the angle, is by definition the cosine of the angle for the same reason. Using the Pythagorean Theorem, the cosine squared is equal to the square of the hypotenuse minus the square of x:

(cos A)^2 = 1^2 - x^2.

Now, we have already said that the sine of the angle is x. So we can substitute (sin A) for x in the above formula:

(cos A)^2 = 1^2 - (sin A)^2.

Algebraic manipulation gets us the proper form:

(sin A)^2 + (cos A)^2 = 1.

For the same angle, if the hypotenuse is multiplied by a scalar, r, then similar triangles demand that every linear piece of the triangle be multiplied by the same scalar. So (cos A) becomes (r cos A) and (sin A) becomes (r sin A). The above formula then becomes:

(r sin A)^2 + (r cos A)^2 = (r)^2
r^2 (sin A)^2 + r^2 (cos A)^2 = r^2

We factor out r^2:

r^2 [(sin A)^2 + (cos A)^2] = r^2.

This implies that [(sin A)^2 + (cos A)^2] = 1, because only r^2 x 1 = r^2.

Answer 3

Remember soh, cah, toa? It's a mnemonic to remember how to calculate sines, cosines, and tangents. Let's think of a right triangle where the angle we're interested in is called A. The side opposite of A is of length a, the adjacent side has length b, and the hypotenuse has length c. Then

sin A = a/c and
cos A = b/c.

So if we look at

sin^2 A + cos^2 A = (a/c)^2 + (b/c)^2
= a^2/c^2 + b^2/c^2
= (a^2+b^2)/c^2

But we said that this is a right triangle with legs of length a and b and hypotenuse of length c. So that means, by the Pythagorean theorem, that a^2 + b^2 = c^2. So

sin^2 A + cos^2 A = (a^2+b^2)/c^2
= c^2/c^2
= 1.

Answer 4

Sure I'll help you out.

Think of a simple right triangle with sides A & B and hypotenuse C

sin = opposite/hypotenuse
cos = adjacent/hypotenuse

using one of the angles, sin = A/C & cos = B/C

plugging that into the equation sin^2 + cos^2 = 1 gives you (A/C)^2 + (B/C)^2 = 1

(A/C)^2 + (B/C)^2 = 1
A^2/C^2 + B^2/C^2 = 1
(A^2 + B^2)/C^2 = 1
multiplying both sides by C^2 gives you
A^2 + B^2 = C^2 (The Pythagorean Theorem)
which has already been proven a million times to be true.

Therefore, sin^2 + cos^2 = 1

Answer 5

o = Opposite
a = Adjacent
h = Hypotenuse

sin(x) = o/h
cos(x) = a/h

sin^2(x) = o^2/h^2
cos^2(x) = a^2/h^2

Pythagoras says that o^2 + a^2 = h^2
Which equals o^2/h^2 + a^2/h2 = 1
Then substitute the above sin^2 and cos^2
sin^2(x) + cos^2(x) = 1